Question

A small ball is projected along the surface of a smooth inclined plane with speed 10 m/s along the direction shown at t = 0. the point of projection is origin, z - axis is along vertical . The accelearation due to gravity is 10 m/ $s^2$. Lists values of certain parameters related to motion of ball and column - II linst fifferent time instants. Match appropriately.

"$$\begin{array}{cc}coloumn1& coloumn2\\ \left(p\right)Distancefromx-axis& \left(a\right)0.5\\ \left(q\right)Speedisminimum& \left(b\right)1.0\\ \left(r\right)Velocitymakesangel{37}^{\circ }withx-axis& \left(c\right)1.50\\ \left(s\right)Ballismaximumawayfromx-axis& \left(d\right)2.0\end{array}$$

• A. p - a,c; q - b; r - d; s - b
• B. p - b; q - a,d; r - d; s - b
• C. p - d; q - b; r - a,b; s - a
• D. p - a; q - c; r - d; s - b

Answer 1

The correct option is A

p - a,c; q - b; r - d; s - b

If we resolve g, we get g cos $\theta$ perpendicular to the plane and g sin $\theta$ along the plane. The perpendicular component is centralized by normal force.

$\therefore$ The ball experience an acceleration of only g sin $\theta$ along the plane.

Now to make it more convienent , we are going to consider only the plane of the surface of the wedge. I am going to call is the x y' plane.

In this plane , the ball has an accelearation of g sin $\theta$ $\frac{m}{s^2}$ in the negative y' direction.

$\therefore$ It is just projectile motion with an acceleration of g sin $\theta$

$\therefore$ x = $\mu$ cos $\theta$ t , y' = $\mu$ sin $\theta$ t - $\frac{1}{2}$ g sin $\theta$ $t^2$

For y = 2.25, $\mu$ = 10 , $\theta$ = ${37}^{\circ }$

2.25 = 10 sin 37 t - $\frac{1}{2}$ * 10 * sin ${37}^{\circ }$ $t^2$